Final answer:
To find the absolute maximum and minimum of f(x) = x² - x + 16 on [0, 12], evaluate f at its critical point and endpoints. The maximum value of 148 occurs at x=12, and the minimum value of 15 + 1/4 occurs at x=1/2.
Step-by-step explanation:
The question is asking to find the absolute maximum and absolute minimum of the function f(x) = x² − x +16 on the interval [0, 12]. To find these values, we need to consider both the critical points of the function within the interval and the values at the endpoints of the interval. To find the critical points, we first calculate the derivative of f(x), which is f'(x) = 2x - 1. Setting the derivative equal to zero gives us the critical point: 2x - 1 = 0 → x = 1/2. Because 1/2 is within the interval [0, 12], we must evaluate f(x) at x = 1/2. Moreover, we evaluate f(x) at the endpoints of the interval, namely x = 0 and x = 12.
Calculating the values:
- f(0) = 0² − 0 + 16 = 16
- f(1/2) = (1/2)² − (1/2) + 16 = 15 + 1/4
- f(12) = 12² − 12 + 16 = 144 − 12 + 16 = 148
Comparing these values, the absolute maximum value of f on the interval [0, 12] is 148 and occurs at x = 12, and the absolute minimum value is 15 + 1/4 and occurs at x = 1/2.